已知tanθ=2√2,θ∈(π/4,π/2),求[2cos²θ/2-sinθ-1]/[√3sin(π/3+θ)sin(π/3-θ)]的值

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√3sin(π/3+θ)sin(π/3-θ)=√3/2{cos[(π/3+θ)-(π/3-θ)]-cos[(π/3+θ)+(π/3-θ)]}
=√3/2{cos[(π/3+θ-π/3+θ)]-cos[(π/3+θ+π/3-θ)]}
=√3/2{cos（2θ)-cos[(2π/3)]}
=√3/2[cos2θ-cos(2π/3)]
=√3/2cos2θ-√3/2cos(2π/3)
=√3/2cos2θ-√3/2cos(120°)
=√3/2cos2θ-√3/2*（-1/2)
=√3/2cos2θ+√3/4
注意cos[(π/3+θ)-(π/3-θ)]-cos[(π/3+θ)+(π/3-θ)]
=cos(π/3+θ)cos(π/3-θ)+sin(π/3+θ)sin(π/3-θ）-[os(π/3+θ)cos(π/3-θ)-in(π/3+θ)sin(π/3-θ)]
=2sin(π/3+θ)sin(π/3-θ)
即sin(π/3+θ)sin(π/3-θ)=1/2{cos[(π/3+θ)-(π/3-θ)]-cos[(π/3+θ)+(π/3-θ)]}请问下√3sin(π/3+θ)sin(π/3-θ)=√3/2{cos[(π/3+θ)-(π/3-θ)]-cos[(π/3+θ)+(π/3-θ)]}这步使用的什么公式积化和差公式sinasinb=1/2[cos（a-b）-cos（a+b）]